The series $\sum_{n = 1}^{\infty} a_{n}$ with $a_{n} = \frac{1}{\sqrt n - i n}$ does not converge absolutely because, $$|a_{n}| = \frac{1}{|\sqrt n - i n|} = \frac{1}{\sqrt{n + n^2}} \geq \frac{1}{\sqrt 2 \ n } $$ and $\frac{1}{\sqrt 2}\sum_{n=1}^{\infty} \frac{1}{n}$ does not converge.

Therefore the ratio test and other comparison tests dont work. My question is, if this series does converge or if it diverges. I dont managed to show if the series is either a cauchy sequence or not

  • $\begingroup$ You are correct. It does not converge absolutely. In fact, it does not converge at all. $\endgroup$ – GEdgar Jun 24 '18 at 21:02
  • $\begingroup$ Im sure there are converging series that dont converge absolutely $\endgroup$ – haddel Jun 24 '18 at 21:07
  • $\begingroup$ $\sum\frac{(-1)^n}{n}$ converges, but not absolutely. $\endgroup$ – GEdgar Jun 24 '18 at 21:09

$$\frac{1}{\sqrt{n}-in} = \frac{\sqrt{n}}{n+n^2}+i\frac{n}{n+n^2} $$ From the definition of a complex series, it doesn't converge, since it's imaginary part doesn't converge. ($\frac{n}{n+n^2} $ is comparable with $\frac{1}{n}$ and $\sum \frac{1}{n} $ doesn't converge)

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  • $\begingroup$ When the absolute value of a complex number (in your case of the denominator of the summand) comes out negative, you can be pretty sure that you have made a mistake. $\endgroup$ – Alex B. Jun 24 '18 at 21:11
  • $\begingroup$ @AlexB. Thank you $\endgroup$ – Jakobian Jun 24 '18 at 22:01

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